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INTEGER-VALUED POLYNOMIALS on KRULL RINGS Introduction If A PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 124, Number 12, December 1996, Pages 3595–3604 S 0002-9939(96)03442-9 INTEGER-VALUED POLYNOMIALS ON KRULL RINGS SOPHIE FRISCH (Communicated by Wolmer V. Vasconcelos) Abstract. If R is a subring of a Krull ring S such that RQ is a valuation ring for every finite index Q = P R, P in Spec1(S), we construct polynomials that map R into the maximal possible∩ (for a monic polynomial of fixed degree) 1 power of PSP, for all P in Spec (S) simultaneously. This gives a direct sum decomposition of Int(R, S), the S-module of polynomials with coefficients in the quotient field of S that map R into S, and a criterion when Int(R, S)has a regular basis (one consisting of 1 polynomial of each non-negative degree). Introduction If A is an infinite subset of a domain S, we write Int(A, S)fortheS-module of polynomials with coefficients in the quotient field of S that – when acting as a function by substitution of the variable – map A into S.ForInt(S, S), the ring of integer-valued polynomials on S, we write Int(S). Beyond the fact (known of x x(x 1)...(x n+1) old) that the binomial polynomials n = − n! − form a basis of the free Z-module Int(Z), the study of Int(S)originatedwithP´olya [16] and Ostrowski [15], who let S be the ring of integers in a number field (their results have been generalized to Dedekind rings by Cahen [4]). Int(R, S)forR=Shas only begun to attract attention more recently [2], [3], [6], [8], [11], [13]. 6 We will treat P´olya’s and Ostrowski’s questions in the case where R = S and S is a Krull ring; in particular the question when Int(R, S)isafreeS-module6 that admits a regular basis, and the related one of determining the highest power of PSP, where P is a height 1 prime ideal of S, that a monic polynomial of fixed degree can map R into. Following P´olya, we call a sequence of polynomials (gn)n 0 regular, ∈N if deg gn = n for all n. One basic connection between a module of polynomials and the modules of leading coefficients should be kept in mind: 0.1 Lemma. Let R be a unitary subring of a field K, M an R-submodule of K[x], and In = leading coefficients of n-th degree polynomials in M 0 . { }∪{ } (i) If (gn)n 0 is a regular sequence of monic polynomials in K[x] such that ∈N Ingn M for all n,thenM= n∞=0 Ingn (direct sum). (ii) A regular⊆ set of polynomials in M is an R-basis if and only if the leading P coefficient of the n-th degree polynomial generates In as an R-module. (iii) M has a regular R-basis if and only if each In is non-zero and cyclic. Received by the editors September 2, 1994 and, in revised form, May 1, 1995. 1991 Mathematics Subject Classification. Primary 13B25, 13F05; Secondary 13F20, 11C08. c 1996 American Mathematical Society 3595 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3596 SOPHIE FRISCH Proof. (i) If (gn)n 0 is as stated, then n∞=0 Ingn M and the sum is direct, since ∈N ⊆ deg(gn)=nmakes the gn linearly independent over K. An induction on N =degf N P shows that f M implies f Ingn. Indeed, for N =0,f I0 =g0I0,and ∈ ∈ n=0 ∈ if N>0andaN is f’s leading coefficient, then aN IN ,soh=f aNgN M N 1 P ∈ − ∈ and h − Ingn by induction hypothesis. (ii) and (iii) are easy. ∈ n=0 1.P Polynomials mapping a set into a discrete valuation ring Throughout section one, v is a discrete valuation on a field K with value-group Γv = Z and v(0) = ,andRv its valuation ring with maximal ideal Mv.Ina kind of generic local regular∞ basis theorem, we will establish the connection (well- known in special cases) between Int(A, Rv) and the maximal power of Mv that a monic polynomial of degree n can map A into, for all A K for which this maximum exists for every n. A subset A of the quotient field⊆ of a domain R is called R-fractional if there exists a d R 0 such that dA R. ∈ \{ } ⊆ 1.0 Lemma. If R is an integrally closed domain with quotient field L, A L and f non-constant L[x] then f(A) is R-fractional if and only if A is. ⊆ ∈ Proof. Let f L[x], deg f = n>0. If f(A)isR-fractional there is a non-zero d R,withdf∈ (a) R for every a A.Letc R 0 , such that cf R[x], ∈ ∈ n ∈ ∈ \{ } ∈ and set g = cdf = cnx + ...+c0. For every a A, g(a) R implies that cna is ∈ ∈ integral over R, therefore cna R and cnA R. The converse is clear. ∈ ⊆ Since a set B K is Rv-fractional if and only if minb B v(b)existsinZ ⊆ ∈ ∪ , Lemma 1.0 shows that A being Rv-fractional is necessary and sufficient for {∞} mina A v(f(a)) to exist in Z for any non-constant f K[x]. To exclude ∈ ∪{∞} ∈ polynomials identically zero on A, for which mina A v(f(a)) = , we need deg f< A, so that the conditions on A in Lemma 1.1 below∈ are necessary.∞ | | 1.1 Lemma. Let n N0.IfAis an Rv-fractional subset of K with A >n,then max min v(f(a)) f∈monic K[x], deg f = n exists. | | {a A | ∈ } ∈ Proof. The case n = 0 is trivial; so let n>0andm Nsuch that A is not m ∈ contained in any union of n cosets of Mv in K. Such an m exists, since n< A m | | and by the Krull Intersection Theorem m Mv = (0). We show that for every ∈N monic f K[x] of degree n there exists an a0 A with v(f(a0)) <nm(and ∈ T ∈ consequently max mina A v(f(a)) f monic K[x], deg f = n <nm). { ∈ | ∈ } Let v0 be an extension of v to the splitting field of f over K, Rv0 its valuation- ring with maximal ideal Mv0 ,ande=[Γv0 :Γv]. A is not contained in any union me me of n cosets of Mv in K0.Pickana0 Athat is not in u + Mv for any root 0 ∈ n 0 u of f in K0;thenv(f(a0)) = v0(f(a0)) = v0(a0 ui) <nm. i=1 − 1.2 Theorem. Let A be an infinite, Rv-fractionalP subset of K.Forn N0set ∈ γv,A(n)=max min v(f(a)) f monic K[x], deg f = n . {a A | ∈ } ∈ γv,A(n) (i) Mv− = leading coefficients of degree n polynomials in Int(A, Rv) 0 . { }∪ { } (ii) A regular basis of Int(A, Rv) is given by (cngn)n 0 ,withgn K[x]monic, ∈N ∈ deg gn = n,andcn K, such that min v(gn(a)) = γv,A(n) and v(cn)= ∈ a A ∈ γv,A(n). − License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER-VALUED POLYNOMIALS 3597 Proof. Let In,v = leading coefficients of degree n polynomials in Int(A, Rv) { }∪ 0 . The leading coefficient cn of any n-th degree polynomial in Int(A, Rv)must { } γv,A(n) satisfy v(cn) γv,A(n), so In,v Mv− .Now,forn N0,letgn be monic ≥− ⊆ ∈ of degree n in K[x] with mina A v(gn(a)) = γv,A(n) (such things exist by dint of ∈ γv,A(n) γv,A(n) Lemma 1.1). Then Mv− gn Int(A, Rv), so Mv− In,v. This shows ⊆ ⊆ (i) and also that In,vgn Int(A, Rv) for all n N0. (ii) follows by Lemma 0.1 and γv,A(n)⊆ ∈ the fact that Mv− = cnRv for every cn K with v(cn)= γv,A(n). ∈ − Before deriving a formula for max mina A v(f(a)) f monic K[x], deg f = n , { ∈ | ∈ } when A is a subring of Rv, we check that the other plausible way of normalizing the polynomials would yield the same value. We also see that polynomials mapping A Rv into the maximal possible power of Mv can be chosen to split with their ⊆ roots in any set that Mv-adically approximates A (for instance in A itself, or, if Rv is the localization of a ring R at a prime ideal of finite index, in R). We need a lemma from [7] (but include the proof). 1.3 Lemma. Let f Rv [x], not all of whose coefficients lie in Mv, split over K, ∈ as f(x)=d(x b1) ... (x bm) (x c1) ... (x cl) with v(bi) < 0,v(ci) 0, − · · − · − · · − ≥ and put f+(x)=(x c1) ... (x cl). Then, for all r Rv, v(f(r)) = v(f+(r)). − · · − ∈ m Proof. For r Rv v(r bi )=v(bi)andsov(f(r)) = v(d)+ i=1 v(bi)+v(f+(r)); ∈ m− 1 n n 1 we show v(d)= j=1 v(bi). Consider d− f(x)=x +an 1x − +...+a0.Since − −P f Rv[x] Mv[x], v(d)= min0 k n v(ak). But ak is the elementary symmetric ∈ \ P − ≤ ≤ polynomial of degree n k in the bi and ci, so the minimal valuation is attained m − by v(an m)= i=1 v(bi).
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